Question

Answer any three que's ._.​

Answer 1

Question :

Draw a neat diagram of a full wave rectifier & explain its working.`

Answer :

Full wave rectifier:

For full wave rectifier, we use two junction diodes. The circuit diagram for a full wave rectifier using two junction diodes is shown in below figure A.

Suppose during the first half cycle of input AC signal the terminal S1 is positive relative to S and S2 is negative relative to S, then diode ∣ is forward biased and diode ∣∣ is reverse biased. Therefore current flows in diode ∣ and not in diode ∣∣. The direction of current i1 due to diode ∣ in load resistance RL is directed from A to B. In the next half cycle, the terminal S1 is negative relative to S and S2 is positive relative to S. Then diode ∣ is reverse biased and diode ∣∣ is forward biased. Therefore current flows in diode ∣∣ and there is no current in diode ∣. The direction of current i2 due to diode ∣∣ in load resistance is again from A to B. Thus for input AC signal the output current is a continuous series of unidirectional pulses. This output current may be converted in fairly steady current by the use of suitable filters. 

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Question :

2) Draw a neat diagram of a rull wave rectimer ex explatinis wo ` Draw input & output characteristics for a full wave rectifier.`

Answer:

Full wave rectifier is a circuit arrangement which makes use of both half cycles of input alternating current (AC) and convert them to direct current (DC). Thus a full wave rectifier is

much more efficient (double+) than a half wave rectifier. This process of converting both half cycles of the input supply (alternating current) to direct current (DC) is termed full wave rectification. Full wave rectifier can be constructed in 2 ways. The first method makes use of a center tapped transformer and 2 diodes.

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Question :

` Describe how a potentiometer is used to compare the two cells by combination method.`

Answer :

For this, the necessary circuit is shown in the figure . Across the ends, A and B of the potentiometer wire are joined a storage cell B1, rheostat Rh, and key K1. The positive pole of B1 is connected to the end A of the wire. Now, the positive poles of the two cells E1 and E2 whose emf's are to be compared are connected to A and the negative poles to the jockey J through a two-way key K2 and a shunted galvanometer G. 

The key K1 and is closed so that a potential difference is established between the ends of the wire AB. Now, by the means of the key K2, the cell E1 is included in the circuit and the null-point is determined by the jockey. Suppose the null-point on the wire is at a length l1 from point A. Then, we have

E1=Kl1

where K is the potential-gradient along the wire, Similarly, the other cells E2 is included in the circuit and again the null-point is determined. Let the length to this null-point from the point A be l2. Then, 

E2=Kl2

E1/E2=l1/l2. 

From this, the ratio E1 and E2 can be calculated. Since the measurements l1 and l2 are taken in the condition of no current, the internal resistances of the sources of emf do not enter the picture. If one of the two cells is a standard cell, whose emf is known, then the emf of the other cell can be determined. 

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